EE 205 Dr. A. Zidouri
Electric Circuits II
The Transformer
Lecture #23
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EE 205 Dr. A. Zidouri
The material to be covered in this lecture is as follows:
o Introduction to Transformers
o Analysis of a Linear Transformer Circuit
o Reflected Impedance
After finishing this lecture you should be able to:
¾
¾
¾
¾
Construct the frequency domain equivalent circuit of a transformer
Analyze a linear transformer circuit
Calculate the reflected impedance into the primary
Calculate the scaling factor for the reflected impedance
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EE 205 Dr. A. Zidouri
Introduction
¾ A Transformer is a magnetic device that takes advantage of the phenomenon of mutual inductance.
¾ A Transformer is generally a four-terminal device comprising two (or more) magnetically coupled
coils.
¾ A Transformer may also be regarded as one whose flux is proportional to the currents in its
windings.
Consider the following general circuit model for a transformer used to connect a load to a source.
jωMI2
jωMI1
ZS
VS
a
I1
R1
jωM
jωL1
b
R2
c
I2
jωL2
d
Fig.33-1 Frequency-domain circuit model for a Transformer
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EE 205 Dr. A. Zidouri
Terminology
¾ The transformer winding connected to the source is the primary winding.
¾ The transformer winding connected to the load is the secondary winding.
Where:
R1 → The resistance of the primary winding
R2 → The resistance of the secondary winding
L1 → The self inductance of the primary winding
L2 → The self inductance of the secondary winding
M → The mutual inductance
VS → The internal Voltage of the sinusoidal source
ZS → The internal impedance of the sinusoidal source
ZL → The Load impedance.
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EE 205 Dr. A. Zidouri
Analysis of a linear transformer circuit
The analysis of the circuit consists of finding I1 and I2 as a function of circuit parameters:
R1, R2, L1, L2, M, VS, ZS, ZL and ω.
Mesh 1
VS = ( ZS +R1 + jωL1 ) I1 - jωMI2
(33-1)
Mesh 2
0 = -jωMI1 + ( ZL +R2 + jωL2 ) I2
(33-2)
For simplification, let
Z11 = R1 + jωL1 + ZS
Z22 = R2 + jωL 2 + ZL
Self impedance of the primary
(33-3)
Self impedance of the secondary (33-4)
Thus,
VS = Z11I1 - jωMI2
0 = -jωMI1 + Z 22I2
From (33-6) yields:
I2 =
(33-5)
(33-6)
jωM
I1
Z 22
(33-7)
Substitute in (33-5) yields
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EE 205 Dr. A. Zidouri
⎛
Z11Z 22 + ω2M2
ω2M2 ⎞
VS =
I1 = ⎜ Z11 +
⎟ I1
Z 22
Z
22 ⎠
⎝
(33-8)
Therefore
Z 22
VS
Z11Z 22 + ω2M2
jωM
jωM
I2 =
VS =
I1
2 2
Z11Z 22 + ω M
Z 22
I1 =
Primary Current
(33-9)
Secondary Current
(33-10)
The impedance at the terminals of the source is:
Z int - Z S
so,
ω2M2
ω2M2
Z ab = Z11 +
− Z S = R1 + jωL1 +
Z 22
R2 + jωL 2 + ZL
Impedance looking into the primary
(33-11)
Note: The impedance Zab is independent of the magnetic polarity of the transformer.
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EE 205 Dr. A. Zidouri
Reflected Impedance
ω2M2
The quantity Zr =
R2 + jωL 2 + ZL
is called the Reflected Impedance
(33-12)
This is due solely to the existence of mutual inductance.
To consider reflected impedance in more detail, we express the load impedance in rectangular form:
ZL = RL + jXL
(33-13)
where XL is positive for inductive load and negative for capacitive load.
Zr in rectangular form will be:
ω2M2 ⎡⎣(RL +R 2 ) - j ( ωL 2 + XL ) ⎤⎦
Zr =
2
2
(RL +R2 ) + ( ωL2 + XL )
Zr =
ω2M2
Z 22
2
⎡⎣(RL +R2 ) - j ( ωL 2 + XL ) ⎤⎦ =
(33-14)
ω2M2
Z 22
2
Z∗22
(33-15)
Therefore the Linear Transformer reflects the conjugate of the self-impedance of the secondary circuit
( Z 22 )
∗
into the primary circuit.
The coefficient
ω2M2
Z 22
2
is called the scaling factor of the reflected impedance.
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EE 205 Dr. A. Zidouri
Example 33-1
Calculate the reflected impedance, the scaling factor, the primary and secondary currents for the Circuit
of Fig.33-2:
Parameters of the transformer
R 2 = 40 Ω,
X1 = 400 Ω,
RL = 360 Ω,
XL = 200 Ω,
ZL
R1 = 100 Ω, X2 = 100 Ω, ωM = 80 Ω,
245
ZS = 184 + j0 Ω, VS =
∠0o V
2
M = k L1L 2
M = 0.4
( 0.125 )( 0.5 ) = 0.1 ⇒ ω = 800 rad/s
Fig.33-2 Circuit for Example 33-1
The reflected impedance:
Substituting equation (33-4) in equation (33-15) we get
Z 22 = R 2 + jωL 2 + ZL = (RL +R2 ) + j ( X2 + XL ) = 400 + j300
Z22 = 400 + j300 = 500Ω and Zr =
ω2M2
Z 22
2
⎡⎣(RL +R2 ) - j ( ωL 2 + XL ) ⎤⎦
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EE 205 Dr. A. Zidouri
therefore
802
Zr =
400 - j300] =1.62 × 4 - j1.62 ×3 = 10.24 - j7.68 Ω
2 [
500
802
64
∗
=
= 0.0256
Z
The scaling factor by which 22 is reflected is
5002 2500
The primary current I1 is given by equation 33-9 as:
Z 22
I1 =
VS
Z11Z 22 + ω2M2
where Z11 = R1 + jωL1 + ZS = 184 +100 + j400 = 284 + j400Ω and Z22 = 400 + j300
400 + j300
245 o
∠0
Therefore I1 =
2
( 284 + j400 )( 400 + j300 ) + 80 2
500∠36.87o
245
∠0o
=
o
245280∠91.53 + 6400 2
0.5
∠ - 53.13o A
2
o
In the time domain, i1 = 0.5 cos 800t - 53.13 A
=
(
)
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EE 205 Dr. A. Zidouri
The secondary current I2 is given by equation 33-10 as:
jωM
I2 =
I1
Z 22
where Z22 = 400 + j300 and ωM = 80
80∠90o 0.5
∠ - 53.13o
Therefore I2 =
o
500∠36.87 2
80
0.08
I2 =
=
A
500 2
2
In the time domain, i2 = 0.08 cos 800t A
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EE 205 Dr. A. Zidouri
Self Test:
For the following circuit of Fig.33-3
200
500+j100 a
100
j1200
c
I1
I2
300
V1
j3600
j1600 V2
b
800-j2500
™ What is the self-impedance of
the primary?
a) 500+j100
b) 700+j3700
c) 200+j3600
d) 1000+j4400
Correct answer b
™ What is the self-impedance of
the secondary?
a) 900-j3700
b) 800-j2500
c) 900-j900
d) 100+j1600
Correct answer c
™ What is the impedance
reflected into the primary?
a) 800-j2500
b) 100+j1600
c) 1000+j4400
d) 800+j800
Correct answer d
d
Fig.33-3 for SelfTest
™ What is the impedance seen looking into the
primary terminals?
a) 700+j4800
b) 200+j3600
c) 1000+j4400
d) 800+j800
Correct answer c
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